If \[y=\frac{\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}-1}}{\sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}-1}}\], then \[\frac{dy}{dx}=\]

\[2x+\frac{{{x}^{3}}}{\sqrt{{{x}^{4}}-1}}\]

\[2x+\frac{2{{x}^{3}}}{\sqrt{{{x}^{4}}-1}}\]

\[x+\frac{2{{x}^{3}}}{\sqrt{{{x}^{4}}-1}}\]

If \[y=\frac{\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}-1}}{\sqrt{{{x}^{2}}+1}

1.	If \[y=\frac{\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}-1}}{\sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}-1}}\], then \[\frac{dy}{dx}=\]
A.	\[2x+\frac{2{{x}^{3}}}{\sqrt{{{x}^{4}}-1}}\]
B.	\[2x+\frac{{{x}^{3}}}{\sqrt{{{x}^{4}}-1}}\]
C.	\[x+\frac{2{{x}^{3}}}{\sqrt{{{x}^{4}}-1}}\]
D.	None of these
Answer» B. \[2x+\frac{{{x}^{3}}}{\sqrt{{{x}^{4}}-1}}\]

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If \[y=\frac{\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}-1}}{\sqrt{{{x}^{2}}+1}-\sqrt{{{x}^{2}}-1}}\], then \[\frac{dy}{dx}=\]

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